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Each question below is followed by two statements I and II. You have to determine whether the data given in the statements are sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer. How many males are there in society A? I. Number of females in society B is 56 more than the number of males in society C. Society C has a total population(male and female combined) of 180. II. Society C has 20% more males than society A who makes 60% of the total population of society A and females in society A is equal to males in society B. 

Each question below is followed by two statements I and II. You have to determine whether the data given in the statements are sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer. How many males are there in society A? I. Number of females in society B is 56 more than the number of males in society C. Society C has a total population(male and female combined) of 180. II. Society C has 20% more males than society A who makes 60% of the total population of society A and females in society A is equal to males in society B.  Correct Answer <p>4) If the data even in both the statements I and II together are not sufficient to answer the question.</p>

Let the number of males in society A be MA

Using Statement I

Let females in society B be FB and Males and females in society C be MC and FC

Then, FB = 56 + MC and MC + FC = 180

So, statement I alone is not sufficient to answer the question

 

Using Statement II

Society C has 20% more males than society A 

⇒ MC = (120/100) × MA = (6/5) × MA

Male population makes 60% of the total population of society A 

If total population of society A be x

MA = (60/100) × x

Then female population in society A

FA = (40/100) × x

Females in society A is equal to males in society B

⇒ FA = MB

So, statement II alone is not sufficient to answer the question

Using statement I and statement II together

MC = (6/5) × (6/10) × x

FB = 56 + MC

∴ No relation can be established

∴ The statements I and II together are not sufficient to answer the question.

 

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